Answering An Open Problem on $T$-Norms for Type-2 Fuzzy Sets

TL;DR

This paper proves that certain convolution-based operations on functions over [0,1] are t-norms or t-conorms, resolving an open problem and characterizing their properties in the context of Type-2 fuzzy sets.

Contribution

It establishes that specific convolution operations are t-norms or t-conorms, answering a previously open question and providing new insights into their characteristics.

Findings

Confirmed convolution operations are t-norms or t-conorms.

Provided characterizations of t-norm and t-conorm via convolution operations.

Resolved an open problem from prior research.

Abstract

This paper proves that a binary operation ${\star}$ on ${[0, 1]}$, ensuring that the binary operation ${\curlywedge}$ is a ${t}$-norm or ${\curlyvee}$ is a ${t}$-conorm, is a ${t}$-norm, where ${\curlywedge}$ and ${\curlyvee}$ are special convolution operations defined by $${(f\curlywedge g)(x)=\sup\left\{f(y)\star g(z): y\vartriangle z=x\right\},} $$ $${(f\curlyvee g)(x)=\sup\left\{f(y)\star g(z): y\ \triangledown\ z=x\right\},} $$ for any ${f, g\in Map([0, 1], [0, 1])}$, where ${\vartriangle}$ and ${\triangledown}$ are a continuous ${t}$-norm and a continuous ${t}$-conorm on ${[0, 1]}$, answering negatively an open problem posed in \cite{HCT2015}. Besides, some characteristics of ${t}$-norm and ${t}$-conorm are obtained in terms of the binary operations ${\curlywedge}$ and ${\curlyvee}$.

Equations

$$(f⋏g)(x)=\sup \left\{f(y)\star g(z):y△z=x\right\},$$

$$(f⋎g)(x)=\sup \left\{f(y)\star g(z):y▽z=x\right\},$$

$${\chi }_B(x)=\{\begin{array}{ll}1,& x\in B,\\ 0,& x\in X\setminus B.\end{array}$$

$$(f\bullet g)(x)=▼\{f(y)▲g(z):y\circ z=x\}.$$

$$(f⋏g)(x)=\sup \left\{f(y)\star g(z):y△z=x\right\},$$

$$(f⋎g)(x)=\sup \left\{f(y)\star g(z):y▽z=x\right\}.$$

$$A:X\to \mathbf{M},$$

$$(f\bigsqcup g)(x)=\sup \{f(y)\wedge g(z):y\vee z=x\},$$

$$(f\sqcap g)(x)=\sup \{f(y)\wedge g(z):y\wedge z=x\},$$

$$(\neg f)(x)=\sup \{f(y):1-y=x\}=f(1-x).$$

$$f^L(x)=\sup \{f(y):y\le x\},$$

$$f^R(x)=\sup \{f(y):y\ge x\}.$$

$$(f⋏g)(1)=f(1)\star g(1).$$

$$(f⋏g)(1)=\sup \{f(y)\star g(z):y△z=1\}=f(1)\star g(1).$$

$$f(x)=(u-1)x+1,$$

$$g(x)=(v-1)x+1,$$

$$u\star v=f(1)\star g(1)=(f⋏g)(1)$$

$$=(g⋏f)(1)=g(1)\star f(1)=v\star u,$$

$$f(x)=(u-1)x+1,$$

$$g(x)=(v-1)x+1,$$

$$h(x)=(w-1)x+1,$$

$$u\star (v\star w)=f(1)\star (g⋏h)(1)=(f⋏(g⋏h))(1)$$

$$=((f⋏g)⋏h)(1)=(f⋏g)(1)\star h(1)=(u\star v)\star w,$$

$${\mathcal{W}}_x(t)=\{\begin{array}{ll}0,& t\in [0,x),\\ t,& t\in [x,1],\end{array}$$

$$0$$

$$={\chi }_{\{1\}}(1)\star {\chi }_{\{0\}}(1)=1\star 0.$$

$$0={\mathcal{W}}_x(t)=\sup \{{\mathcal{W}}_x(y)\star {\chi }_{\{1\}}(z):y△z=t\}.$$

$$0\ge {\mathcal{W}}_x(x)\star {\chi }_{\{1\}}(z_1)=x\star 0,\text{ i.e., }x\star 0=0.$$

$$0\ge {\chi }_{\{0\}}(y)\star {\chi }_{\{1\}}(\frac{1}{2})=0\star 0,\text{ i.e., }0=0\star 0.$$

$$x\star 0=0\star x=0.$$

$$1$$

$$={\chi }_{\{1\}}(1)\star {\chi }_{\{1\}}(1)=1\star 1.$$

Full text

[TABLE]

[TABLE]

for any ${f,g\in Map([0,1],[0,1])}$, where ${\vartriangle}$ and ${\triangledown}$ are a continuous ${t}$-norm and a continuous ${t}$-conorm on ${[0,1]}$, answering negatively an open problem posed in [17]. Besides, some characteristics of ${t}$-norm and ${t}$-conorm are obtained in terms of the binary operations ${\curlywedge}$ and ${\curlyvee}$.

This work was supported by the National Natural Science Foundation of China (No. 11601449), the Science and Technology Innovation Team of Education Department of Sichuan for Dynamical Systems and its Applications (No. 18TD0013), and the Youth Science and Technology Innovation Team of Southwest Petroleum University for Nonlinear Systems (No. 2017CXTD02).

1. Introduction

In 1975, Zadeh [1] introduced the notion of type-2 fuzzy sets (T2FSs) – that is, fuzzy set with fuzzy sets as truth values (simply, “fuzzy-fuzzy sets”) – being an extension of type-1 fuzzy sets (FSs) and interval-valued fuzzy sets (IVFSs), which was also equivalently expressed in different forms by Mendel et al. ([2]–[5]). Because the truth values of T2FSs are fuzzy, they are more adaptable to a further study of uncertainty than FSs and have been applied in many studies ([6]–[25]). Mendel [6] summarized some important advances for FSs and T2FSs from 2001 to 2007. Hu and Kwong [7] discussed $t$-norm operations of T2FSs and obtained a few properties of type-2 fuzzy numbers. For better understanding of T2FSs, Aisbett et al. [8] translates their constructs to the language of functions in spaces. Chen and Wang [9] used T2FSs to give a new technique for fuzzy multiple attributes decision making. Sola et al. [10] provided a more general perspective for interval T2FSs and showed that IVFSs can be viewed as a special case of interval T2FSs. Ruiz et al. [11] obtained two results for join and meet operations for T2FSs with arbitrary secondary memberships. Recently, Wang  [12] introduced the notion of conditional fuzzy sets to characterize T2FSs. Then, Wu et al. [13] presented a Jaccard similarity measure for general T2FSs, as an extension of the Jaccard similarity measure for FSs and IVFSs.

Being an extension of the logic connective conjunction and disjunctionin classical two-valued logic, triangular norms ($t$-norms) with the neutral $1$ and triangular conorms ($t$-conorms) with the neutral [math] on the unit interval $I=[0,1]$ were introduced by Menger [14] in 1942 and by Schweizer and Sklar [15] in 1961, respectively. Because $t$-norms and $t$-conorms have a close connection with fuzzy set theory and order related theories, they play an important role in many fields, such as fuzzy set theory [26], fuzzy logic [16], fuzzy systems modeling [27], and probabilistic metric spaces [15]. Walker and Walker [28] extended $t$-norms and $t$-conorms to the algebra of truth values of T2FSs. Then, Hernándes et al. [17] introduced the notions of $t_{r}$-norm and $t_{r}$-conorm by adding some “restrictive axioms” (see Definition 2 below) with systematic analysis. In particular, they [17] proved that the following binary operation $\curlywedge$ (resp., $\curlyvee$) on the set of all normal and convex functions constructed by convolution is a $t_{r}$-norm (resp., a $t_{r}$-conorm). Recently, we proved [29] that the fuzzy metric $M$ of every stationary fuzzy metric space $(X,M,\star)$ is uniformly continuous.

Throughout this paper, let $I=[0,1]$, $Map(X,Y)$ be the set of all mappings from $X$ to $Y$, and ‘$\leq$’ denote the usual order relation in the lattice of real numbers. In particular, let $\mathbf{M}=Map(I,I)$ and $\mathbf{L}$ be the set of all normal and convex functions in $\mathbf{M}$.

Definition 1**.**

[24] A $t$-norm on $I$ is a binary operation $\star:I^{2}\rightarrow I$ satisfying

1. (T1)

(commutativity/symmetry) $x\star y=y\star x$ for $x,y\in I$; 2. (T2)

(associativity) $(x\star y)\star z=x\star(y\star z)$ for $x,y,z\in I$; 3. (T3)

(increasing) $\star$ is increasing in each argument; 4. (T4)

(neutral element) $1\star x=x\star 1=x$ for $x\in I$.

A binary operation $\star:I^{2}\rightarrow I$ is a $t$-conorm on $I$ if it satisfies axioms (T1), (T2), and (T3) above; axiom (T4’): $0\star x=x\star 0=x$ for $x\in I$.

For any subset $B$ of $X$, a special fuzzy set $\chi_{B}$, which is called the characteristic function of $B$, is defined as

[TABLE]

Definition 2**.**

[17] A binary operation $T:\mathbf{L}^{2}\rightarrow\mathbf{L}$ is a $t_{r}$-norm ($t$-norm according to the restrictive axioms), if

1. (O1)

$T$ is commutative, i.e., $T(f,g)=T(g,f)$ for $f,g\in\mathbf{L}$; 2. (O2)

$T$ is associative, i.e., $T(T(f,g),h)=T(f,T(g,h))$ for $f,g,h\in\mathbf{L}$; 3. (O3)

$T(f,\chi_{\{1\}})=f$ for $f\in\mathbf{L}$ (neutral element); 4. (O4)

letting $f,g,h\in\mathbf{L}$ such that $g\sqsubseteq h$; then, $T(f,g)\sqsubseteq T(f,h)$ (increasing in each argument); 5. (O5)

$T(\chi_{[0,1]},\chi_{[a,b]})=\chi_{[0,b]}$; 6. (O6)

$T$ is closed on $\mathbf{J}$; 7. (O7)

$T$ is closed on $\mathbf{K}$;

where $\mathbf{J}$ is the set of all characteristic functions of the elements of $I$, and $\mathbf{K}$ is the set of all characteristic functions of the closed subintervals of $I$, i.e., $\mathbf{J}=\{\chi_{\{x\}}:x\in I\}$, $\mathbf{K}=\{\chi_{[a,b]}:0\leq a\leq b\leq 1\}$.

A binary operation $S:\mathbf{L}^{2}\rightarrow\mathbf{L}$ is a $t_{r}$-conorm if it satisfies axioms (O1), (O2), (O4), (O6), and (O7) above; axiom (O3’): $S(f,\chi_{\{0\}})=f$; and axiom (O5’): $S(\chi_{[0,1]},\chi_{[a,b]})=\chi_{[a,1]}$. Axioms (O1), (O2), (O3), (O3’), and (O4) are called “basic axioms”, and an operation that complies with these axioms will be referred to as $t$-norm or $t$-conorm, respectively.

Convolution as a standard way to combine functions was used to construct operations on $Map(J,[0,1])$. Let $\circ$ and $\blacktriangle$ be two binary operations defined on $X$ and $Y$, respectively, and $\blacktriangledown$ be an appropriate operation on $Y$. Define a binary operation $\bullet$ on the set $Map(X,Y)$ by

[TABLE]

This method of defining an operation on $Map(X,Y)$ from operations on $X$ and $Y$ is called convolution.

Definition 3**.**

[17] Let $\star$ be a binary operation on $I$, $\vartriangle$ be a $t$-norm on $I$, and $\triangledown$ be a $t$-conorm on $I$. Define the binary operations $\curlywedge$ and $\curlyvee:\mathbf{M}^{2}\rightarrow\mathbf{M}$ as follows: for $f,g\in\mathbf{M}$,

[TABLE]

and

[TABLE]

In 2015, Hernándes et al. [17] proposed the following open problem on the binary operations $\curlywedge$ and $\curlyvee$.

Question 4**.**

[17] Apart from the $t$-norms, does there exist other binary operation ‘$\star$’ on $I$ such that ‘$\curlywedge$’ and ‘$\curlyvee$’ are, respectively, a $t_{r}$-norm and a $t_{r}$-conorm on $\mathbf{L}$?

This paper first gives a negative answer to Question 4, proving that, if a binary operation $\star$ ensures that $\curlywedge$ is a $t_{r}$-norm on $\mathbf{L}$ or $\curlyvee$ is a $t_{r}$-conorm on $\mathbf{L}$, then $\star$ is a $t$-norm, i.e., $\star$ satisfies axioms (T1)–(T4). Then, it is proved that the following are equivalent:

1. (1)

$\star$ is a $t$-norm on $I$; 2. (2)

$\curlywedge$ is a $t_{r}$-norm on $\mathbf{L}$; 3. (3)

$\curlywedge$ is a $t$-norm on $\mathbf{L}$; 4. (4)

$\curlyvee$ is a $t_{r}$-conorm on $\mathbf{L}$; 5. (5)

$\curlyvee$ is a $t$-conorm on $\mathbf{L}$.

Finally, analogous results on $\vartriangle$ are presented when the binary operation $\star$ is restricted to be a continuous $t$-norm.

2. Preliminaries

A type-1 fuzzy set $A$ in space $X$ is a mapping from $X$ to $I$, i.e., $A\in Map(X,I)$, and $A(x)$ is called the degree of membership of an element $x\in X$ to the set $A$. The two sets $\O$ and $X$ are special elements in $Map(X,I)$, with $\O(x)\equiv 0$ and $X(x)\equiv 1$, respectively. A fuzzy set $A\in Map(X,I)$ is normal if $\sup\{A(x):x\in I\}=1$.

Definition 5**.**

[17] A function $f\in\mathbf{M}$ is convex if, for any $x\leq y\leq z$, it holds that $f(y)\geq f(x)\wedge f(z)$.

Definition 6**.**

[30] A type-2 fuzzy set $A$ in space $X$ is a mapping

[TABLE]

i.e., $A\in Map(X,\mathbf{M})$. For any $x\in X$, $A(x)$ is also called the degree of membership of an element $x\in X$ to the set $A$.

Definition 7**.**

[30] The operations of $\sqcup$ (union), $\sqcap$ (intersection), $\neg$ (complementation) on $\mathbf{M}$ are defined as follows: for any $f,g\in\mathbf{M}$,

[TABLE]

[TABLE]

and

[TABLE]

From [30], it follows that $\mathbb{M}=(\mathbf{M},\sqcup,\sqcap,\neg,\chi_{\{0\}},\chi_{\{1\}})$ does not have a lattice structure, although $\sqcup$ and $\sqcap$ satisfy the De Morgan’s laws with respect to the given operation $\neg$.

Walker and Walker [30] introduced the following partial order on $\mathbf{M}$.

Definition 8**.**

[30] $f\sqsubseteq g$ if $f\sqcap g=f$; $f\preceq g$ if $f\sqcup g=g$.

It follows from [30, Proposition 14] that both $\sqsubseteq$ and $\preceq$ are partial orders on $\mathbf{M}$. In [22, 23, 30], it was proved that the subalgebra $\mathbb{L}=(\mathbf{L},\sqcup,\sqcap,\neg,\chi_{\{0\}},\chi_{\{1\}})$ is a bounded complete lattice. In particular, $\chi_{\{0\}}$ and $\chi_{\{1\}}$ are the minimum and maximum, respectively.

For $f\in\mathbf{M}$, define $f^{L}$ and $f^{R}$ in $\mathbf{M}$ by

[TABLE]

and

[TABLE]

Clearly, $f^{L}$ and $f^{R}$ are monotonically increasing and decreasing, respectively. The following properties of $f^{L}$ and $f^{R}$ are obtained by Walker et al. ([22, 23, 30]).

Proposition 9**.**

[30] For $f,g\in\mathbf{M}$,

1. (1)

$f\sqsubseteq g$ if and only if $f^{R}\wedge g\leq f\leq g^{R}$; 2. (2)

$f\preceq g$ if and only if $f\wedge g^{L}\leq g\leq f^{L}$; 3. (3)

$f\leq f^{L}$, $f\leq f^{R}$; 4. (4)

$(f^{L})^{L}=f^{L}$, $(f^{R})^{R}=f^{R}$; 5. (5)

$(f^{L})^{R}=(f^{R})^{L}=\sup f$;

Theorem 10**.**

([22, 23]) Let $f,g\in\mathbf{L}$. Then, $f\sqsubseteq g$ if and only if $g^{L}\leq f^{L}$ and $f^{R}\leq g^{R}$.

Lemma 11**.**

For $f\in\mathbf{M}$, $f^{L}(0)=f(0)$ and $f^{R}(1)=f(1)$.

Proof.

From the definitions of $f^{L}$ and $f^{R}$, this holds trivially. ∎

3. Answer to the Open Problem

3.1. Commutativity and Associativity of $\star$

Lemma 12**.**

Let $\star$ be a $t$-norm on $I$. Then, $x\star y=1$ if and only if $x=y=1$.

Lemma 13**.**

Let $\vartriangle$ be a continuous $t$-norm on $I$ and $\star$ be a binary operation on $I$. Then,

[TABLE]

Proof.

Since $\vartriangle$ is a $t$-norm, from Lemma 12, it follows that

[TABLE]

∎

Proposition 14**.**

Let $\vartriangle$ be a continuous $t$-norm on $I$ and $\star$ be a binary operation on $I$. Then,

1. (1)

$\curlywedge$ and $\curlyvee$ are commutative on $\mathbf{L}$ if and only if $\star$ is commutative; 2. (2)

If $\curlywedge$ and $\curlyvee$ are associative on $\mathbf{L}$, then $\star$ is associative.

Proof.

(1) The sufficiency follows from the proof of [17, Proposition 1]. It remains to prove the necessity. Suppose on the contrary that $\star$ is not commutative. Then, there exist $u,v\in I$ such that $u\star v\neq v\star u$. Choose two functions $f,g\in\mathbf{M}$ as

[TABLE]

and

[TABLE]

for $x\in I$. It can be verified that $f,g\in\mathbf{L}$, as $f$ and $g$ are decreasing. Since $\curlywedge$ is commutative, applying Lemma 13 yields that

[TABLE]

which is a contradiction. Therefore, $\star$ is commutative.

(2) Suppose on the contrary that $\star$ is not associative. Then, there exist $u,v,w\in I$ such that $u\star(v\star w)\neq(u\star v)\star w$. Choose three functions $f,g,h\in\mathbf{M}$ as

[TABLE]

[TABLE]

and

[TABLE]

for $x\in I$. It can be verified that $u,v,w\in\mathbf{L}$, as $f$, $g$, and $h$ are decreasing. Since $\curlywedge$ is associative, applying Lemma 13 yields that

[TABLE]

which is a contradiction. Therefore, $\star$ is associative. ∎

Remark 15*.*

Similar results to Proposition 14 are obtained by Hernández et al. [17] under the assumption that $\curlywedge$ and $\curlyvee$ are commutative or associative on $\mathbf{M}$, which is stronger than the condition in Proposition 14.

3.2. Neutral Element $1$ for $\star$

For any fixed $x\in I$, define $\mathscr{W}_{x}:I\rightarrow I$ by

[TABLE]

for $t\in I$. It can be verified that $\mathscr{W}_{x}\in\mathbf{L}$, as $\mathscr{W}_{x}$ is increasing for $x\in I$.

Lemma 16**.**

Let $\vartriangle$ be a continuous $t$-norm on $I$ and $\star$ be a binary operation on $I$. If $\curlywedge$ is a $t$-norm on $\mathbf{L}$, then $0\star x=x\star 0=0$ for all $x\in I$.

Proof.

(1) As $\chi_{\{1\}}$ is a neural element, by Lemma 13, one has

[TABLE]

(2) Fix any $x\in(0,1)$. From $\mathscr{W}_{x}(t)=(\mathscr{W}_{x}\curlywedge\chi_{\{1\}})(t)=\sup\{\mathscr{W}_{x}(y)\star\chi_{\{1\}}(z):y\vartriangle z=t\}$, it follows that, for any $t\in(0,x)$,

[TABLE]

Since \vartriangle$$(x,-) is continuous on $[0,1]$, and \vartriangle$$(x,0)=0, \vartriangle$$(x,1)=x, it follows from the intermediate value theorem that there exists some $z_{1}\in(0,1)$ such that \vartriangle$$(x,z_{1})=x$$\vartriangle$$z_{1}=t. This, together with (3.1), implies that

[TABLE]

(3) Note that $0=\chi_{\{0\}}(\frac{1}{4})=(\chi_{\{0\}}\curlywedge\chi_{\{1\}})(\frac{1}{4})=\sup\left\{\chi_{\{0\}}(y)\star\chi_{\{1\}}(z):y\vartriangle z=\frac{1}{4}\right\}$. Similarly to the proof of (2), it follows that there exists $y\in(0,1)$ such that $y\vartriangle\frac{1}{2}=\frac{1}{4}$. This implies that

[TABLE]

Summing up (1)–(3) and the commutativity of $\star$ (Proposition 14), it follows that, for any $x\in[0,1]$,

[TABLE]

∎

Lemma 17**.**

Let $\vartriangle$ be a continuous $t$-norm on $I$ and $\star$ be a binary operation on $I$. If $\curlywedge$ is a $t$-norm on $\mathbf{L}$, then $1\star x=x\star 1=x$ for all $x\in I$.

Proof.

(1) Since $\chi_{\{1\}}$ is a neural element, from Lemma 13, it follows that

[TABLE]

(2) For any fixed $x\in(0,1)$, $x=\mathscr{W}_{x}(x)=(\mathscr{W}_{x}\curlywedge\chi_{\{1\}})(x)=\sup\left\{\mathscr{W}_{x}(y)\star\chi_{\{1\}}(z):y\vartriangle z=x\right\}$. For $y,z\in I$ with $y\vartriangle z=x$, consider the following two cases:

Case 1. If $z=1$, then $y=x$. This implies that $\mathscr{W}_{x}(y)\star\chi_{\{1\}}(z)=x\star 1$;

Case 2. If $z<1$, then $\chi_{\{1\}}(z)=0$. Applying Lemma 16 gives that

[TABLE]

Thus,

[TABLE]

The proof is completed by applying $0\star 1=0$ and the commutativity of $\star$. ∎

3.3. Increasing in Each Argument for $\star$

For any fixed $x\in I$, define $\mathscr{V}_{x}:I\rightarrow I$ by

[TABLE]

It can be verified that $\mathscr{V}_{x}\in\mathbf{L}$, as $\mathscr{V}_{x}$ is decreasing for $x\in I$. Clearly, functions $f$, $g$, and $h$ constructed in Proposition 14 satisfy that $f=\mathscr{V}_{u}$, $g=\mathscr{V}_{v}$, and $h=\mathscr{V}_{w}$.

Applying the decreasing property of $\mathscr{V}_{x}$ immediately yields the following result.

Lemma 18**.**

For any $x\in I$, $\mathscr{V}^{L}_{x}\equiv 1$ and $\mathscr{V}^{R}_{x}=\mathscr{V}_{x}$.

Lemma 19**.**

For any $x_{1},x_{2}\in I$ with $x_{1}\leq x_{2}$, $\mathscr{V}_{x_{1}}\sqsubseteq\mathscr{V}_{x_{2}}$.

Proof.

Clearly, $\mathscr{V}_{x_{1}}\leq\mathscr{V}_{x_{2}}$. Applying Lemma 18 yields that

[TABLE]

and

[TABLE]

This, together with Theorem 10, implies that

[TABLE]

∎

Lemma 20**.**

Let $\vartriangle$ be a continuous $t$-norm on $I$ and $\star$ be a binary operation on $I$. If $\curlywedge$ is a $t$-norm on $\mathbf{L}$, then for any $y\in(0,1)$ the functions $\star^{r}_{y}$ and $\star^{l}_{y}$ are increasing, where $\star^{r}_{y}(x)=x\star y$ and $\star^{l}_{y}(x)=y\star x$ for $x\in I$.

Proof.

It follows from Proposition 14 that $\star^{r}_{y}=\star^{l}_{y}$. So, it suffices to prove that $\star^{r}_{y}$ is increasing.

For any $0\leq x_{1}\leq x_{2}\leq 1$, since $\curlywedge$ is increasing in each argument, from Lemma 19, it follows that

[TABLE]

In particular, by Theorem 10,

[TABLE]

This, together with Lemmas 11 and 13, implies that

[TABLE]

i.e.,

[TABLE]

Therefore, $\star^{r}_{y}$ is increasing. ∎

3.4. Answer to Question 4

Theorem 21**.**

Let $\vartriangle$ be a continuous $t$-norm on $I$ and $\star$ be a binary operation on $I$. If $\curlywedge$ is a $t$-norm on $\mathbf{L}$, then $\star$ is a $t$-norm.

Proof.

It follows directly from Proposition 14, and Lemmas 17 and 20. ∎

Similarly, the following result can be verified.

Theorem 22**.**

Let $\triangledown$ be a continuous $t$-conorm on $I$ and $\star$ be a binary operation on $I$. If $\curlyvee$ is a $t$-conorm on $\mathbf{L}$, then $\star$ is a $t$-norm.

Remark 23*.*

Theorems 21 and 22 show that a binary operation $\star$ on $I$, ensuring that $\curlywedge$ is a $t$-norm (thus a $t_{r}$-norm) on $\mathbf{L}$ or $\curlyvee$ is a $t$-conorm (thus a $t_{r}$-conorm) on $\mathbf{L}$, must be a $t$-norm. This give a negative answer to Question 4.

Combining together Theorems 21, 22, and [17, Proposition 14], one obtains the following.

Theorem 24**.**

Let $\vartriangle$ be a continuous $t$-norm, $\triangledown$ be a continuous $t$-conorm, and $\star$ be a continuous binary operation on $I$. Then, the following statements are equivalent:

1. (1)

$\star$* is a $t$-norm on $I$;* 2. (2)

$\curlywedge$* is a $t_{r}$-norm on $\mathbf{L}$;* 3. (3)

$\curlywedge$* is a $t$-norm on $\mathbf{L}$;* 4. (4)

$\curlyvee$* is a $t_{r}$-conorm on $\mathbf{L}$;* 5. (5)

$\curlyvee$* is a $t$-conorm on $\mathbf{L}$.*

4. Further Study on the Binary Operation $\curlywedge$

Let $\star$ be a continuous $t$-norm on $I$ and $\vartriangle$ be a surjective binary operation on $I$. Define the binary operation $\curlywedge:\mathbf{M}^{2}\rightarrow\mathbf{M}$ as follows: for $f,g\in\mathbf{M}$,

[TABLE]

Here, the surjection assumption on $\vartriangle$ is necessary, because $(f\curlywedge g)(x)$ is not well defined for every point $x$ in $I\setminus\vartriangle(I^{2})$, if $\vartriangle$ is not surjective. Motivated by Question 4, a fundamental question is: Apart from the $t$-norms, does there exist other binary operation ‘$\vartriangle$’ on $I$ such that ‘$\curlywedge$’ is a $t_{r}$-norm on $\mathbf{L}$?

This section will also give a negative answer to this question.

Lemma 25**.**

For $x_{1},x_{2}\in I$, $\chi_{\{x_{1}\}}\sqsubseteq\chi_{\{x_{2}\}}$ if and only if $x_{1}\leq x_{2}$.

Proof.

Firstly, it can be verified that, for any $x\in I$,

[TABLE]

and

[TABLE]

This, together with Theorem 10, implies that

[TABLE]

∎

Lemma 26**.**

Let $\star$ be a continuous $t$-norm on $I$ and $\vartriangle$ be a binary operation on $I$. Then, for any $x_{1},x_{2}\in I$, $\chi_{\{x_{1}\}}\curlywedge\chi_{\{x_{2}\}}=\chi_{\{x_{1}\vartriangle x_{2}\}}$.

Proof.

Since $\star$ is a continuous $t$-norm, applying Lemma 12 gives

1. (a)

for $y,z\in I$, $\chi_{\{x_{1}\}}(y)\star\chi_{\{x_{2}\}}(z)\in\{0,1\}$; 2. (b)

$\chi_{\{x_{1}\}}(y)\star\chi_{\{x_{2}\}}(z)=1$ if and only if $y=x_{1}$ and $z=x_{2}$.

This, together with $(\chi_{\{x_{1}\}}\curlywedge\chi_{\{x_{2}\}})(x)=\sup\{\chi_{\{x_{1}\}}(y)\star\chi_{\{x_{2}\}}(z):y\vartriangle z=x\}$, implies that

[TABLE]

∎

Lemma 27**.**

Let $\star$ be a continuous $t$-norm on $I$ and $\vartriangle$ be a binary operation on $I$. Then,

1. (1)

$\curlywedge$* is commutative on $\mathbf{L}$ if and only if $\vartriangle$ is commutative;* 2. (2)

If $\curlywedge$ is associative on $\mathbf{L}$, then $\vartriangle$ is associative.

Proof.

(1) The sufficiency holds trivially. It remains to check the necessity.

For $x,y\in I$, since $\curlywedge$ is commutative, it follows from Lemma 26 that

[TABLE]

implying that

[TABLE]

Thus, $\vartriangle$ is commutative.

(2) For $x,y,z\in I$, since $\curlywedge$ is associative, it follows from Lemma 26 that

[TABLE]

implying that

[TABLE]

Thus, $\vartriangle$ is associative. ∎

Lemma 28**.**

Let $\star$ be a continuous $t$-norm on $I$ and $\vartriangle$ be a binary operation on $I$. If $\curlywedge$ is a $t$-norm on $\mathbf{L}$, then $1\vartriangle x=x\vartriangle 1=x$ for all $x\in I$.

Proof.

For any $x\in I$, since $\chi_{\{1\}}$ is an neutral element, applying Lemma 26 yields that

[TABLE]

Thus, $1\vartriangle x=x=x\vartriangle 1$. ∎

Lemma 29**.**

Let $\star$ be a continuous $t$-norm on $I$ and $\vartriangle$ be a binary operation on $I$. If $\curlywedge$ is a $t$-norm on $\mathbf{L}$, then, for any $y\in(0,1)$, the functions $\vartriangle^{r}_{y}$ and $\vartriangle^{l}_{y}$ is increasing, where $\vartriangle^{r}_{y}(x)=x\vartriangle y$ and $\vartriangle^{l}_{y}(x)=y\vartriangle x$ for any $x\in I$.

Proof.

It follows from Lemma 27 that $\vartriangle^{r}_{y}=\vartriangle^{l}_{y}$. So, it suffices to prove that $\vartriangle^{r}_{y}$ is increasing.

For $0\leq x_{1}\leq x_{2}\leq 1$, applying Lemma 25 follows that

[TABLE]

Since $\curlywedge$ is increasing in each argument, applying Lemma 26 yields that

[TABLE]

This, together with Lemma 25, implies that

[TABLE]

Therefore, $\vartriangle^{R}_{y}$ is increasing. ∎

Combining together Lemmas 27–29 and [17, Proposition 14] immediately yields the following result.

Theorem 30**.**

Let $\star$ be a continuous $t$-norm on $I$ and $\vartriangle$ be a continuous binary operation on $I$. Then, the following statements are equivalent:

1. (1)

$\vartriangle$* is a $t$-norm on $I$;* 2. (2)

$\curlywedge$* is a $t_{r}$-norm on $\mathbf{L}$;* 3. (3)

$\curlywedge$* is a $t$-norm on $\mathbf{L}$.*

Similarly, one can obtain the following result.

Theorem 31**.**

Let $\star$ be a continuous $t$-norm on $I$ and $\vartriangle$ be a continuous binary operation on $I$. Then, the following statements are equivalent:

1. (1)

$\vartriangle$* is a $t$-conorm on $I$;* 2. (2)

$\curlywedge$* is a $t_{r}$-conorm on $\mathbf{L}$;* 3. (3)

$\curlywedge$* is a $t$-conorm on $\mathbf{L}$.*

5. Conclusion

This paper has further studied the binary operations $\curlywedge$ and $\curlyvee$ defined in (1.1), (1.2) and (4.1) on $\mathbf{L}$. By introducing two special families of functions $\mathscr{W}_{x}$ and $\mathscr{V}_{x}$ ($x\in I$), it first proves that, when the continuous $t$-norm $\vartriangle$ or continuous $t$-conorm $\triangledown$ is fixed, the following hold:

1. (1)

$\curlywedge$ is a continuous $t_{r}$-norm on $\mathbf{L}$ if and only if $\curlywedge$ is a continuous $t$-norm on $\mathbf{L}$ if and only if $\star$ is a continuous $t$-norm; 2. (2)

$\curlyvee$ is a continuous $t_{r}$-conorm on $\mathbf{L}$ if and only if $\curlyvee$ is a continuous $t$-conorm on $\mathbf{L}$ if and only if $\star$ is a continuous $t$-norm.

In particular, these results negatively answer Question 4. Similarly to Question 4, the case that the binary operation $\vartriangle$ is fixed (see (4.1)) has been investigated and some analogous results were obtained.